Thomas Simpson and Maxima and Minima - Motion of bodies I

Two bodies move at the same time, from two given places A and B, and proceed uniformly from thence in given directions, AP and BQ, with celerities in a given ratio; it is proposed to find their position, and how far each has gone, when they are nearest possible to each other (Example XIII, page 28).

Let a = AC, b = BC, c = DC in the adjacent figure. Assume m is the velocity ("celerity") of the body that moves in the direction AP while n is the velocity of the body that moves in the direction BN.

At time t the first body will be at M while the second body will be at N (Simpson calls these two points "cotemporary"). Next draw perpendiculars NE and BD to AP, and let x = CN. Since DECN is similar to DDCB, we can conclude that b/x = c/CE. Because M and N are cotemporary points it follows that AM/m = BN/n, therefore AM = m/n(x-b). So CM = AC-AM = a-m/n(x-b) = d-m/nx, where d = a + m/nb. By the law of cosines we have MN2 = CM2 + CN2- 2(CM)(CN)cosC. But cosC = EC/CN. Therefore

MN2

= CM2 + CN2- 2(CM)(CE) =

æ
è

d-

m

n

x

ö
ø

2

+ x2- 2

æ
è

d-

m

n

x

ö
ø

cx

b

= d2-

æ
è

2dm

n

+

2cd

b

ö
ø

x +

æ
è

m2

n2

+

2cm

nb

+1

ö
ø

x2.

Taking the derivative of MN2, which is obviously a function of x, and making it equal to zero we get

x =

mnbd + cdn2

bm2 + 2cmn + bn2

.

This is the value where MN2, and consequently MN, attains its minimum.

Remark: Maybe Simpson should have mentioned that MN2 is a second degree polynomial in the variable x, thus on the Cartesian plane it represents a vertical parabola that opens upwards with minimum at -r/2s where r and s are the coefficients of x and x2 respectively. We would get exactly the same value obtained before, namely

x =

mnbd + cdn2

bm2 + 2cmn + bn2

.

So, in this problem there is an alternative to the use of derivatives.